\(2xy-x^2-y^2+16\)

\(=-\left(x^2-2xy+y^2-16\right)\)

\(=-\left(x-y-4\right)\left(x-y+4\right)\)

a: \(4x^2-4x\)

\(=4x\cdot x-4x\cdot1\)

\(=4x\left(x-1\right)\)

b: \(x^2-2xy+y^2-4\)

\(=\left(x-y\right)^2-2^2\)

\(=\left(x-y-2\right)\left(x-y+2\right)\)

`4x^2-4x`

`= 4x(x-1)`

__

`x^2 -2xy +y^2-4`

`= (x^2-2xy+y^2)-2^2`

`=(x-y)^2 -2^2`

`=(x-y-2)(x-y+2)`

a,

\(4x^2-4x\\=4x(x-1)\)

b,

\(x^2-2xy+y^2-4\\=(x^2-2xy+y^2)-4\\=(x-y)^2-2^2\\=(x-y-2)(x-y+2)\)

\(\text{#}Toru\)

a) \(3\left(x-y\right)^2+9y\left(y-x\right)^2\)

\(=3\left(x-y\right)^2+9y\left(x-y\right)^2\)

\(=\left(x-y\right)^2\left(3-9y\right)\)

\(=3\left(x-y\right)^2\left(3y+1\right)\)

b) \(3\left(x-y\right)^2+9y\left(y-x\right)\)

\(=3\left(y-x\right)^2+9y\left(y-x\right)\)

\(=\left(y-x\right)\left[3\left(y-x\right)+9y\right]\)

\(=3\left(y-x\right)\left(y-x+3y\right)\)

\(=3\left(y-x\right)\left(4y-x\right)\)

a: =3(x-y)^2+9y(x-y)^2

=(x-y)^2(3+9y)

=(x-y)^2*3*(y+3)

b: =3(x-y)^2-9y(x-y)

=3(x-y)(x-y-9y)

=3(x-y)(x-10y)

a: =(x-z)(y+8)

b; =x^2-2x-3x+6

=(x-2)(x-3)

c: =x^4+10x^2-x^2-10

=(x^2+10)(x^2-1)

=(x^2+10)(x-1)(x+1)

-3x⁴y + 6x³y - 3x²y

= -3x²y( x² - 2x + 1 )

= -3x²y( x - 1 )²

-3x⁴y+6x³y-3x²y

=-3x²y(x²-2xy+1)

=-3x²y(x-1)²

x³ - 9x² + 6x + 16

= x³ - 8x² -x² + 8x - 2x + 16

= x²(x-8) -x(x-8) -2(x-8)

= (x-8)(x²-x-2)

= (x-8)(x²-2x + x - 2)

=(x-8)[x(x-2)+(x-2)]

=(x-8)(x-2)(x+1)

a: Ta có: \(25x^2\left(x-y\right)-x+y\)

\(=\left(x-y\right)\left(25x^2-1\right)\)

\(=\left(x-y\right)\left(5x-1\right)\left(5x+1\right)\)

b: Ta có: \(16x^2\left(z^2-y^2\right)-z^2+y^2\)

\(=\left(z^2-y^2\right)\left(16x^2-1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left(4x-1\right)\left(4x+1\right)\)

c: Ta có: \(x^3+x^2y-x^2z-xyz\)

\(=x^2\left(x+y\right)-xz\left(x+y\right)\)

\(=x\left(x+y\right)\left(x-z\right)\)

\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)